AMC 8 · 2020 · #24

Grade 7 geometry-2dalgebra

Archetype Arithmetic Expression Decomposition

area-rectanglespercentageratio-proportionlinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectanglespercentagelinear-equations-one-var

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A large square region is paved with $n^2$ gray square tiles, each measuring $s$ inches on a side. A border $d$ inches wide surrounds each tile. The figure below shows the case for $n=3$ . When $n=24$ , the $576$ gray tiles cover $64\%$ of the area of the large square region. What is the ratio $\frac{d}{s}$ for this larger value of $n?$

$\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}$

Pick an answer.

(A)

$frac{6}{25}$

(B)

$frac{1}{4}$

(C)

$frac{9}{25}$

(D)

$frac{7}{16}$

(E)

$frac{9}{16}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A big square is tiled with $n \times n$ gray squares, each of side $s$ inches, separated and bordered by strips of width $d$ inches. The picture in the problem shows the layout for $n=3$. For $n=24$ the $576$ gray tiles together cover $64\%$ of the big square's area. Find the ratio $\dfrac{d}{s}$.

Givens: Tile arrangement: an $n \times n$ grid of gray squares of side $s$; Borders of width $d$ run between adjacent tiles AND along the outer edge; For $n = 24$ there are $24^2 = 576$ gray tiles; Gray area $= 64\%$ of the big-square area; Answer choices: (A) $\tfrac{6}{25}$, (B) $\tfrac{1}{4}$, (C) $\tfrac{9}{25}$, (D) $\tfrac{7}{16}$, (E) $\tfrac{9}{16}$

Unknowns: The ratio $\dfrac{d}{s}$ of border width to tile side

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #13 Convert to Algebra

$n = 24$ is too big to count borders by hand, so start with Tool #9 and use the $n = 3$ picture the problem already gives. Counting tiles vs. borders in that small case exposes the pattern "$n$ tiles, $n+1$ borders" along one side; Tool #1 (the diagram is already supplied) makes the $n+1$ borders obvious — one between each pair, plus one at each end. Once the side length is $ns + (n+1)d$, Tool #13 turns the percent condition into a single linear equation in $d$ and $s$ that we can solve for the ratio.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.OA.C.5 Step 1

Look at the $n=3$ picture and count along one side.
There are $3$ gray tiles in a row and the borders show up at $4$ places: one before tile 1, one between tiles 1 and 2, one between tiles 2 and 3, and one after tile 3.
So along one side of the big square the length is $3s + 4d$.

$$\text{side}(n{=}3) = 3s + 4d$$

💡 Counting items vs. gaps in the small case is a Grade 4 "find the rule for a pattern" move.

#9 Solve an Easier Related Problem 4.OA.C.5 Step 2

Generalize the count.
For any $n$, a row has $n$ tiles and $n+1$ border strips (one between every pair of adjacent tiles plus one at each end).
Plugging in $n = 24$ gives $24$ tiles and $25$ borders along each side of the big square.

$$\text{side}(n) = n\,s + (n+1)\,d \;\Rightarrow\; \text{side}(24) = 24s + 25d$$

💡 Extending the $n=3$ rule to $n=24$ is exactly the "generate a pattern from a rule" idea.

#1 Draw a Diagram 6.G.A.1 Step 3

Write each square's area.
The gray region is itself a $24s \times 24s$ square (because the $576 = 24^2$ tiles of side $s$ stack into a perfect $24$-by-$24$ block when you ignore the borders), so its area is $(24s)^2$.
The big square has side $24s + 25d$, so its area is $(24s + 25d)^2$.

$$A_{\text{gray}} = (24s)^2,\quad A_{\text{big}} = (24s + 25d)^2$$

💡 Decomposing the big square into a $24 \times 24$ tile block plus surrounding border strips is Grade 6 area-by-composition.

#13 Convert to Algebra 7.G.A.1 Step 4

Translate the $64\%$ condition into the side-length ratio.
Both regions are squares, so the area ratio is the square of the side ratio.
Since $64\% = \left(\tfrac{8}{10}\right)^2 = \left(\tfrac{4}{5}\right)^2$, the gray side is $\tfrac{4}{5}$ of the big side.

$$\dfrac{A_{\text{gray}}}{A_{\text{big}}} = \dfrac{64}{100} = \left(\dfrac{4}{5}\right)^2 \;\Rightarrow\; \dfrac{24s}{24s + 25d} = \dfrac{4}{5}$$

💡 When two figures are similar (here both squares), areas scale with the square of the side ratio — a Grade 7 scale-drawing idea.

#13 Convert to Algebra 6.EE.B.7 Step 5

Cross-multiply and solve the resulting linear equation for the ratio $\dfrac{d}{s}$.
Divide by $s$ at the very end so the answer comes out as a pure ratio independent of the tile size.

$$5 \cdot 24s = 4(24s + 25d) \;\Rightarrow\; 120s = 96s + 100d \;\Rightarrow\; 24s = 100d \;\Rightarrow\; \dfrac{d}{s} = \dfrac{24}{100} = \dfrac{6}{25} \;\Rightarrow\; \textbf{(A)}$$

💡 Solving a single-variable linear equation like $24s = 100d$ for the ratio is the Grade 6 "$px = q$" idea.

[1] #9 4.OA.C.5 Look at the $n=3$ picture and count along one side. There are $3$ gray tiles in

[2] #9 4.OA.C.5 Generalize the count. For any $n$, a row has $n$ tiles and $n+1$ border strips (

[3] #1 6.G.A.1 Write each square's area. The gray region is itself a $24s \times 24s$ square (b

[4] #13 7.G.A.1 Translate the $64\%$ condition into the side-length ratio. Both regions are squa

[5] #13 6.EE.B.7 Cross-multiply and solve the resulting linear equation for the ratio $\dfrac{d}{

Review

Reasonableness: Sanity-check the side ratio directly: the gray block is $24s$ wide and the big square is $24s + 25d$ wide, so the gray fraction (per side) should be $\tfrac{4}{5} = 0.8$. With $d/s = 6/25$, one side is $24s + 25 \cdot \tfrac{6}{25}s = 24s + 6s = 30s$, and $\tfrac{24s}{30s} = \tfrac{4}{5}$. Squaring gives $\tfrac{16}{25} = 0.64 = 64\%$ — exactly the given coverage, so $\boxed{\tfrac{6}{25}}$ is consistent.

Alternative: Tool #3 (Eliminate Possibilities) using the answer choices: for each candidate $d/s$, compute $\tfrac{24}{24 + 25(d/s)}$ and square it; we need $0.64$. Choice (A) gives $\tfrac{24}{24 + 6} = \tfrac{4}{5}$, squared $= 0.64$. Choice (B) gives $\tfrac{24}{24 + 25/4} \approx 0.793$, squared $\approx 0.629$. Choice (C) gives $\tfrac{24}{24 + 9} = \tfrac{24}{33} \approx 0.727$, squared $\approx 0.529$. Only (A) hits $0.64$.

CCSS standards used (min grade 7)

4.OA.C.5 Generate a number or shape pattern following a given rule (Counting $n$ tiles vs. $n+1$ borders in the $n=3$ picture and extending the rule to $n=24$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the big square into a $24 \times 24$ gray block surrounded by border strips so both areas can be written as $(\text{side})^2$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Cross-multiplying $\tfrac{24s}{24s+25d} = \tfrac{4}{5}$ down to $24s = 100d$ and reading off $d/s = 6/25$.)
7.G.A.1 Solve problems involving scale drawings of geometric figures (Using the fact that similar figures' areas scale by the square of the side ratio to convert $64\% = (4/5)^2$ into a side-length ratio.)

⭐ This AMC 8 problem only needs Grade 7 scale-drawing reasoning — areas of similar squares scale by the square of the side ratio — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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